What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?

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19-06-2018
Chemistry

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What is the molality of a solution made by dissolving 15.20 g of i2 in 1.33 mol of diethyl ether, (ch3ch2)2o?

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kenmyna kenmyna · Answer 1 · 2018-06-28T12:40:50+02:00

The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M

calculation

molarity =moles of solute/ Kg of the solvent

mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol

moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles

calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol

mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg

molarity is therefore = 0.06/0.09842 = 0.6 M

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-14T12:57:59+02:00

The molality of a solution made by dissolving 15.20 g of I₂ in 1.33 mol of diethyl ether is 0.6m.

What is molality?

Molality of any substance is define as the number of moles of solute present in per kilograms of the solvent.

Moles of solute from the mass will be calculated as:

n = W/M, where

W = given mass of iodine (I₂) = 15.2g

M = molar mass of iodine (I₂) = 253.8 g/mol

Moles of iodine = 15.2 g/253.8 g/mol = 0.06 moles

Similarly mass of the solvent (CH₃CH₂)₂O will also be calculated by using the abive formula as:

W = 1.33 moles x 74 g/mol = 98.42 grams = 0.09842 Kg

Now molality will be calculated by using all values in the desired formula as:
m = 0.06mol/0.09842kg = 0.6 m

Hence resultant molality of the solution is 0.6m.

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